On attitude polarization under Bayesian learning with non-additive beliefs

Abstract

Ample psychological evidence suggests that people’s learning behavior is often prone to a “myside bias” or “irrational belief persistence” in contrast to learning behavior exclusively based on objective data. In the context of Bayesian learning such a bias may result in diverging posterior beliefs and attitude polarization even if agents receive identical information. Such patterns cannot be explained by the standard model of rational Bayesian learning that implies convergent beliefs. Based on Choquet expected utility theory, we therefore develop formal models of Bayesian learning with psychological bias as alternatives to rational Bayesian learning. We derive conditions under which beliefs may diverge in the learning process despite the fact that all agents observe the same sample drawn from an i.i.d. process. Key to our approach is the description of ambiguous beliefs as neo-additive capacities (Chateauneuf et al., J Econ Theory 137:538–567, 2007), which allows for a flexible and parsimonious parametrization of departures from additive probability measures.

Appendix

1.1 Formal proofs

Proof of Observation 2

By an argument in Schmeidler (1986, p. 256f), it suffices to restrict attention to a non-negative valued random variable Y so that

$$ E\left[ Y,\nu \right] =\int_{0}^{+\infty }\nu \left\{ \omega \in \Omega \mid Y\left( \omega \right) \geq z\right\} dz\text{,} $$

which is equivalent to

$$ E\left[ Y,\nu \right] =\int_{\min Y}^{\max Y}\nu \left\{ \omega \in \Omega \mid Y\left( \omega \right) \geq z\right\} dz \label{shortCE} $$

(22)

since Y is closed and bounded. We consider a partition P _n , n = 1,2,... , of Ω with members

$$ A_{n}^{k}=\left\{ \omega \in \Omega \mid a_{k,n}<X\left( \omega \right) \leq b_{k,n}\right\} \text{ for }k=1,...,2^{n} $$

such that

$$ \begin{array}{rcl} a_{k,n} &=&\left[ \max Y-\min Y\right] \cdot \frac{\left( k-1\right) }{2^{n}} +\min Y \\ b_{k,n} &=&\left[ \max Y-\min Y\right] \cdot \frac{k}{2^{n}}+\min Y\text{.} \end{array} $$

Define the step functions a _n :Ω→ℝ and b _n :Ω→ℝ such that, for $\omega \in A_{n}^{k}$, k = 1,...,2ⁿ,

$$ \begin{array}{rcl} a_{n}\left( \omega \right) &=&a_{k,n} \\ b_{n}\left( \omega \right) &=&b_{k,n}\text{.} \end{array} $$

Obviously,

$$ E\left[ a_{n},\nu \right] \leq E\left[ Y,\nu \right] \leq E\left[ b_{n},\nu \right] $$

for all n and

$$ \lim_{n\rightarrow \infty }E\left[ b_{n},\nu \right] -E\left[ a_{n},\nu \right] =0\text{.} $$

That is, $E\left[ a_{n},\nu \right]$ and $E\left[ b_{n},\nu \right]$ converge to $E\left[ Y,\nu \right]$ for n→ ∞. Furthermore, observe that

$$ \begin{array}{rcl} \min a_{n} &=&\min Y\text{ for all }n\text{, and} \\ \max b_{n} &=&\max Y\text{ for all }n\text{.} \end{array} $$

Since lim_n→ ∞ min b _n  = lim_n→ ∞ min a _n and $E\left[ b_{n},\mu \right]$ is continuous in n, we have

$$ \begin{array}{rcl} \lim_{n\rightarrow \infty }E\left[ b_{n},\nu \right] \!&=&\!\delta \left( \lambda \lim_{n\rightarrow \infty }\max b_{n}\!+\!\left( 1\!-\!\lambda \right) \lim_{n\rightarrow \infty }\min b_{n}\right) \!+\!\left( 1\!-\!\delta \right) \lim_{n\rightarrow \infty }E\left[ b_{n},\mu \right] \\ &=&\delta \left( \lambda \max Y+\left( 1-\lambda \right) \min Y\right) +\left( 1-\delta \right) E\left[ Y,\mu \right] \text{.} \end{array} $$

In order to prove Proposition 3, it therefore remains to be shown that, for all n,

$$ E\left[ b_{n},\nu \right] =\delta \left( \lambda \max b_{n}+\left( 1-\lambda \right) \min b_{n}\right) +\left( 1-\delta \right) E\left[ b_{n},\mu \right] \text{.} $$

Since b _n is a step function, Eq. 22 becomes

$$ \begin{array}{rcl} E\left[ b_{n},\nu \right] &=&\sum\limits_{A_{n}^{k}\in P_{n}}\nu \left( A_{n}^{2^{n}}\cup ...\cup A_{n}^{k}\right) \cdot \left( b_{k,n}-b_{k-1,n}\right) \\ &=&\sum\limits_{A_{n}^{k}\in P_{n}}b_{k,n}\cdot \left[ \nu \left( A_{n}^{2^{n}}\cup ...\cup A_{n}^{k}\right) -\nu \left( A_{n}^{2^{n}}\cup ...\cup A_{n}^{k-1}\right) \right] \text{,} \end{array} $$

implying for a neo-additive capacity

$$ \begin{array}{rcl} E\left[ b_{n},\nu \right] &=&\max b_{n}\left[ \delta \lambda +\left( 1-\delta \right) \mu \left( A_{n}^{2^{n}}\right) \right] + \sum\limits_{k=2}^{2^{n}-1}b_{k,n}\left( 1-\delta \right) \mu \left( A_{n}^{k}\right) \\ &&+\min b_{n}\left[ 1-\delta \lambda -\left( 1-\delta \right) \sum\limits_{k=2}^{2^{n}}\mu \left( A_{n}^{k}\right) \right] \\ &=&\delta \lambda \max b_{n}+\left( 1-\delta \right) \sum\limits_{k=1}^{2^{n}}b_{k,n}\mu \left( A_{n}^{k}\right) +\min b_{n}\left[ \delta -\delta \lambda \right] \\ &=&\delta \left( \lambda \max b_{n}+\left( 1-\lambda \right) \min b_{n}\right) +\left( 1-\delta \right) E\left[ b_{n},\mu \right] \text{.} \end{array} $$

□

Proof of Observation 3

An application of the full Bayesian update rule to a neo-additive capacity gives

$$ \begin{array}{rcl} \nu ^{FB}\left( A\mid B\right) &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) +1-\left( \delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cup \lnot B\right) \right) } \\ &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) }{1+\left( 1-\delta \right) \cdot \left( \mu \left( A\cap B\right) -\mu \left( A\cup \lnot B\right) \right) } \\ &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) }{1+\left( 1-\delta \right) \cdot \left( \mu \left( A\cap B\right) -\mu \left( A\right) -\mu \left( \lnot B\right) +\mu \left( A\cap \lnot B\right) \right) } \\ &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) }{1+\left( 1-\delta \right) \cdot \left( -\mu \left( \lnot B\right) \right) } \\ &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) }{\delta +\left( 1-\delta \right) \cdot \mu \left( B\right) } \\ &=&\frac{\delta \cdot \lambda }{\delta +\left( 1-\delta \right) \cdot \mu \left( B\right) }+\frac{\left( 1-\delta \right) \cdot \mu \left( B\right) }{ \delta +\left( 1-\delta \right) \cdot \mu \left( B\right) }\mu \left( A\mid B\right) \\ &=&\delta _{B}^{FB}\cdot \lambda +\left( 1-\delta _{B}^{FB}\right) \cdot \mu \left( A\mid B\right) \end{array} $$

with $\delta _{B}^{FB}$ given by Eq. 10. □

Proof of Observation 4

An application of the optimistic Bayesian update rule to a neo-additive capacity gives

$$ \begin{array}{rcl} \nu ^{opt}\left( A\mid B\right) &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cap B\right) }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( B\right) } \\ &=&\frac{\delta \cdot \lambda }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( B\right) }+\frac{\left( 1-\delta \right) \cdot \mu \left( B\right) }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( B\right) }\cdot \mu \left( A\mid B\right) \\[4pt] &=&\delta _{B}^{opt}+\left( 1-\delta _{B}^{opt}\right) \cdot \mu \left( A\mid B\right) \end{array} $$

whereby

$$ \delta _{B}^{opt}=\frac{\delta \cdot \lambda }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( B\right) }\text{.} $$

□

Proof of Observation 5

An application of the pessimistic Bayesian update rule to a neo-additive capacity gives

$$ \begin{array}{rcl} \nu ^{pess}\left( A\mid B\right) &=&\frac{\nu \left( A\cup \lnot B\right) -\nu \left( \lnot B\right) }{1-\nu \left( \lnot B\right) } \\[3pt] &=&\frac{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( A\cup \lnot B\right) -\delta \cdot \lambda -\left( 1-\delta \right) \cdot \mu \left( \lnot B\right) }{1-\delta \cdot \lambda -\left( 1-\delta \right) \cdot \mu \left( \lnot B\right) } \\[3pt] &=&\frac{\left( 1-\delta \right) \cdot \mu \left( \lnot \left( \lnot A\cap B\right) \right) -\left( 1-\delta \right) \cdot \mu \left( \lnot B\right) }{ 1-\delta \cdot \lambda -\left( 1-\delta \right) \cdot \mu \left( \lnot B\right) } \\[3pt] &=&\frac{\left( 1-\delta \right) \cdot \left( 1-\mu \left( \lnot A\cap B\right) \right) -\left( 1-\delta \right) \cdot \left( 1-\mu \left( B\right) \right) }{1-\delta \cdot \lambda -\left( 1-\delta \right) \cdot \mu \left( \lnot B\right) } \\[3pt] &=&\frac{\left( 1-\delta \right) \cdot \left( \mu \left( B\right) -\mu \left( \lnot A\cap B\right) \right) }{1-\delta \cdot \lambda -\left( 1-\delta \right) \cdot \left( \mu \left( \lnot B\right) \right) } \\[3pt] &=&\frac{\left( 1-\delta \right) \cdot \left( \mu \left( B\right) -\mu \left( B\right) \mu \left( \lnot A\mid B\right) \right) }{1-\delta \cdot \lambda -\left( 1-\delta \right) \cdot \left( \mu \left( \lnot B\right) \right) } \\[3pt] &=&\frac{\left( 1-\delta \right) \cdot \left( \mu \left( B\right) -\mu \left( B\right) \left( 1-\mu \left( A\mid B\right) \right) \right) }{ 1-\delta \cdot \lambda -\left( 1-\delta \right) \cdot \mu \left( \lnot B\right) } \\[3pt] &=&\left( 1-\delta _{B}^{pess}\right) \cdot \mu \left( A\mid B\right) \end{array} $$

whereby

$$ \delta _{B}^{pess}=\frac{\delta \left( 1-\lambda \right) }{\delta \left( 1-\lambda \right) +\left( 1-\delta \right) \cdot \mu \left( B\right) }\text{. } $$

□

Proof of Observation 6

This observation is an immediate consequence of the following lemma combined with the fact that the additive part of the neo-additive beliefs converges in probability to the true probability π ^∗, i.e.,

$$ \lim_{n\rightarrow \infty }prob\left( \left\vert E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\mid \mathbf{I}_{n}^{k}\right) \right] -\pi ^{\ast }\right\vert \leq c\right) =1 $$

for some c > 0. The lemma simply uses our results (Observations 3–5) on Bayesian updating of neo-additive capacities in order to derive conditional neo-additive capacities for the special case Eq. 17.

Lemma

Suppose that the agent receives sample information $\mathbf{I}_{n}^{k}$. Contingent on the applied update rule we obtain the following conditional neo-additive beliefs and posterior estimates about parameter π whereby $E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\mid \mathbf{I}_{n}^{k}\right) \right]$ is given by Eq. 1.

• (i) Full Bayesian updating.

  $$ \nu ^{FB}\left( \mathbf{\pi }\mid \mathbf{I}_{n}^{k}\right) =\delta _{ \mathbf{I}_{n}^{k}}^{FB}\lambda +\left( 1-\delta _{\mathbf{I} _{n}^{k}}^{FB}\right) \cdot K_{\alpha +k,\beta +n-k}\pi ^{\alpha +k-1}\left( 1-\pi \right) ^{\beta +n-k-1} $$

  with

  $$ \delta _{\mathbf{I}_{n}^{k}}^{FB}=\frac{\delta }{\delta +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}_{n}^{k}\right) } $$

  so that

  $$ E\left[ \tilde{\pi},\nu ^{FB}\left( \mathbf{\cdot }\mid \mathbf{I} _{n}^{k}\right) \right] =\delta _{\mathbf{I}_{n}^{k}}^{FB}\lambda +\left( 1-\delta _{\mathbf{I}_{n}^{k}}^{FB}\right) \cdot E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\mid \mathbf{I}_{n}^{k}\right) \right] . $$

• (ii) Optimistic Bayesian updating.

  $$ \nu ^{opt}\left( \mathbf{\pi }\mid \mathbf{I}_{n}^{k}\right) =\delta _{ \mathbf{I}_{n}^{k}}^{opt}+\left( 1-\delta _{\mathbf{I}_{n}^{k}}^{opt}\right) \cdot K_{\alpha +k,\beta +n-k}\pi ^{\alpha +k-1}\left( 1-\pi \right) ^{\beta +n-k-1} $$

  with

  $$ \delta _{\mathbf{I}_{n}^{k}}^{opt}=\frac{\delta \cdot \lambda }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}_{n}^{k}\right) } $$

  so that

  $$ E\left[ \tilde{\pi},\nu ^{opt}\left( \mathbf{\cdot }\mid \mathbf{I} _{n}^{k}\right) \right] =\delta _{\mathbf{I}_{n}^{k}}^{opt}+\left( 1-\delta _{\mathbf{I}_{n}^{k}}^{opt}\right) \cdot E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\mid \mathbf{I}_{n}^{k}\right) \right] . $$

• (iii) Pessimistic Bayesian updating.

  $$ \nu ^{pess}\left( \mathbf{\pi }\mid \mathbf{I}_{n}^{k}\right) =\left( 1-\delta _{\mathbf{I}_{n}^{k}}^{pess}\right) \cdot K_{\alpha +k,\beta +n-k}\pi ^{\alpha +k-1}\left( 1-\pi \right) ^{\beta +n-k-1} $$

  with

  $$ \delta _{\mathbf{I}_{n}^{k}}^{pess}=\frac{\delta \cdot \left( 1-\lambda \right) }{\delta \cdot \left( 1-\lambda \right) +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}_{n}^{k}\right) } $$

  so that

  $$ E\left[ \tilde{\pi},\nu ^{pess}\left( \mathbf{\cdot }\mid \mathbf{I} _{n}^{k}\right) \right] =\left( 1-\delta _{\mathbf{I}_{n}^{k}}^{pess}\right) \cdot E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\mid \mathbf{I} _{n}^{k}\right) \right] . $$

□

Proof of Observation 7

At first observe that δ > 0 and $\lambda \in \left( 0,1\right)$ implies $\delta _{\mathbf{I} ^{\ast }}^{FB}>\delta _{\mathbf{I}^{\ast }}^{opt}$ as well as $\delta _{ \mathbf{I}^{\ast }}^{FB}>\delta _{\mathbf{I}^{\ast }}^{pess}$. Consider the inequality

$$ \begin{array}{rcl} E\left[ \tilde{\pi},\nu ^{FB}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] &<&E\left[ \tilde{\pi},\nu ^{opt}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] \Leftrightarrow \\ \delta _{\mathbf{I}^{\ast }}^{FB}\cdot \lambda &<&\delta _{\mathbf{I}^{\ast }}^{opt}+\left( \delta _{\mathbf{I}^{\ast }}^{FB}-\delta _{\mathbf{I}^{\ast }}^{opt}\right) \cdot \pi ^{\ast }\text{,} \end{array} $$

which holds, by $\delta _{\mathbf{I}^{\ast }}^{FB}>\delta _{\mathbf{I}^{\ast }}^{opt}$, for all π ^∗ iff

$$ \begin{array}{rcl} \delta _{\mathbf{I}^{\ast }}^{FB}\cdot \lambda &<&\delta _{\mathbf{I}^{\ast }}^{opt}\Leftrightarrow \\ \frac{\delta \cdot \lambda }{\delta +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}^{\ast }\right) } &<&\frac{\delta \cdot \lambda }{\delta \cdot \lambda +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}^{\ast }\right) }\Leftrightarrow \\ \lambda &<&1\text{.} \end{array} $$

Turn now to the inequality

$$ \begin{array}{rcl} E\left[ \tilde{\pi},\nu ^{pess}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] &<&E\left[ \tilde{\pi},\nu ^{FB}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] \Leftrightarrow \\ \left( \delta _{\mathbf{I}^{\ast }}^{FB}-\delta _{\mathbf{I}^{\ast }}^{pess}\right) \cdot \pi ^{\ast } &<&\delta _{\mathbf{I}^{\ast }}^{FB}\cdot \lambda \text{,} \end{array} $$

which holds, by $\delta _{\mathbf{I}^{\ast }}^{FB}>\delta _{\mathbf{I}^{\ast }}^{pess}$, for all π ^∗ iff

$$ \begin{array}{rcl} \left( \delta _{\mathbf{I}^{\ast }}^{FB}-\delta _{\mathbf{I}^{\ast }}^{pess}\right) &<&\delta _{\mathbf{I}^{\ast }}^{FB}\cdot \lambda \Leftrightarrow \\ \delta _{\mathbf{I}^{\ast }}^{FB}\cdot \left( 1-\lambda \right) &<&\delta _{ \mathbf{I}^{\ast }}^{pess}\Leftrightarrow \\ \frac{\delta \cdot \left( 1-\lambda \right) }{\delta +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}^{\ast }\right) } &<&\frac{\delta \cdot \left( 1-\lambda \right) }{\delta \cdot \left( 1-\lambda \right) +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I}^{\ast }\right) }\Leftrightarrow \\ 0 &<&\lambda \text{.} \end{array} $$

This proves the first part of the observation. The second part readily follows from the assumption that $\mu \left( \mathbf{I}^{\ast }\right) <1$. □

Proof of Proposition 1

Part (i). :

Observe at first that inequality (19) is satisfied if and only if λ ₁ ≥ λ ₂. Obviously, if λ ₁ = λ ₂ then Eq. 18 must be violated. Thus we can restrict attention to λ ₁ > λ ₂. Observe that, by the corollary, Eq. 18 writes as

$$ \begin{array}{rcl} &&\delta _{\mathbf{I}^{\ast }}^{FB}\lambda _{1}+\left( 1-\delta _{\mathbf{I} ^{\ast }}^{FB}\right) \cdot \pi ^{\ast }-\delta _{\mathbf{I}^{\ast }}^{FB}\lambda _{2}+\left( 1-\delta _{\mathbf{I}^{\ast }}^{FB}\right) \cdot \pi ^{\ast } \\ &>&\delta \cdot \lambda _{1}+\left( 1-\delta \right) \cdot E\left[ \tilde{\pi },\mu \left( \mathbf{\cdot }\right) \right] -\left( \delta \cdot \lambda _{2}+\left( 1-\delta \right) \cdot E\left[ \tilde{\pi},\mu \left( \mathbf{ \cdot }\right) \right] \right) \text{,} \end{array} $$

which is equivalent to

$$ \begin{array}{rcl} \delta _{\mathbf{I}^{\ast }}^{FB} &>&\delta \Leftrightarrow \\ \frac{\delta }{\delta +\left( 1-\delta \right) \cdot \mu \left( \mathbf{I} ^{\ast }\right) } &>&\delta \text{,} \end{array} $$

and therefore holds if and only if $\delta \in \left( 0,1\right)$ since $\mu \left( \mathbf{I}^{\ast }\right) <1$.

Part (ii). :

Again, observe at first that inequality (19) is satisfied if and only if λ ₁ ≥ λ ₂. By the corollary, Eq. 18 becomes

$$ \begin{array}{rcl} &&\delta _{\mathbf{I}^{\ast }}^{opt}+\left( 1-\delta _{\mathbf{I}^{\ast }}^{opt}\right) \cdot \pi ^{\ast }-\left( 1-\delta _{\mathbf{I}^{\ast }}^{pess}\right) \cdot \pi ^{\ast } \\ &>&\delta \cdot \lambda _{1}+\left( 1-\delta \right) \cdot E\left[ \tilde{\pi },\mu \left( \mathbf{\cdot }\right) \right] -\left( \delta \cdot \lambda _{2}+\left( 1-\delta \right) \cdot E\left[ \tilde{\pi},\mu \left( \mathbf{ \cdot }\right) \right] \right) \end{array} $$

which is equivalent to

$$ \delta _{\mathbf{I}^{\ast }}^{opt}+\left( \delta _{\mathbf{I}^{\ast }}^{pess}-\delta _{\mathbf{I}^{\ast }}^{opt}\right) \cdot \pi ^{\ast }>\delta \left( \lambda _{1}-\lambda _{2}\right) . \label{inequ} $$

(23)

If δ = 0, the l.h.s. as well as the r.h.s. of Eq. 23 equal zero. Thus, δ > 0 is a necessary condition for Eq. 23 to hold. In what follows we prove that δ > 0 is also a sufficient condition. Let δ > 0 and consider at first the case that

$$ \delta _{\mathbf{I}^{\ast }}^{pess}-\delta _{\mathbf{I}^{\ast }}^{opt}\leq 0 \text{.} \label{case1} $$

(24)

Since the l.h.s. of Eq. 23 is then continuously strictly decreasing in π ^∗ and, by assumption, $\pi ^{\ast }\in \left( 0,1\right)$, Eq. 23 is satisfied for all π ^∗ if and only if

$$ \begin{array}{rl} \delta _{\mathbf{I}^{\ast }}^{pess}& \geq \delta \left( \lambda _{1}-\lambda _{2}\right) \Leftrightarrow \\ \frac{1-\lambda _{2}}{\delta (1-\lambda _{2})+(1-\delta )\mu (\mathbf{I} ^{\ast })}& \geq \lambda _{1}-\lambda _{2}\text{,} \end{array} $$

which is obviously true for all λ ₁,λ ₂ since

$$ \frac{1-\lambda _{2}}{\delta (1-\lambda _{2})+(1-\delta )\mu (\mathbf{I} ^{\ast })}\geq 1-\lambda _{2}\text{.} $$

This proves the claim for case (24).

Let δ > 0 and consider now the converse case

$$ \delta _{\mathbf{I}^{\ast }}^{pess}-\delta _{\mathbf{I}^{\ast }}^{opt}>0 \text{.} \label{case2} $$

(25)

Since the l.h.s. of Eq. 23 is then continuously strictly increasing in $\pi ^{\ast }\in \left( 0,1\right)$, Eq. 23 is satisfied for all π ^∗ if and only if

$$ \begin{array}{rcl} \delta _{\mathbf{I}^{\ast }}^{opt}+\left( \delta _{\mathbf{I}^{\ast }}^{pess}-\delta _{\mathbf{I}^{\ast }}^{opt}\right) &\geq &\delta \left( \lambda _{1}-\lambda _{2}\right) \Leftrightarrow \\[3pt] \delta _{\mathbf{I}^{\ast }}^{pess} &\geq &\delta \left( \lambda _{1}-\lambda _{2}\right) \Leftrightarrow \\[3pt] \frac{1-\lambda _{2}}{\delta (1-\lambda _{2})+(1-\delta )\mu (\mathbf{I} ^{\ast })} &\geq &\lambda _{1}-\lambda _{2}\text{,} \end{array} $$

which is obviously true for all λ ₁,λ ₂. This proves that δ > 0 is sufficient for Eq. 23 to hold. □

Proof of Proposition 2

Part (i). :

By the corollary and the assumption that $E\left[ \tilde{ \pi},\mu \left( \mathbf{\cdot }\right) \right] =\pi ^{\ast }$ Eq. 20 implies

$$ \begin{array}{rl} E\left[ \tilde{\pi},\nu _{1}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] &> E\left[ \tilde{\pi},\nu _{2}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] \Leftrightarrow \\ \delta _{\mathbf{I}^{\ast }}^{FB}\lambda _{1}+\left( 1-\delta _{\mathbf{I} ^{\ast }}^{FB}\right) \cdot \pi ^{\ast } & >\delta _{\mathbf{I}^{\ast }}^{FB}\lambda _{2}+\left( 1-\delta _{\mathbf{I}^{\ast }}^{FB}\right) \cdot \pi ^{\ast } \end{array} $$

which holds if and only if λ ₁ > λ ₂ so that the middle inequality in Eq. 20 is also strict. Focus now on the inequalities

$$ \begin{array}{rl} E\left[ \tilde{\pi},\nu _{1}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] &>E\left[ \tilde{\pi},\nu _{1}\left( \mathbf{\cdot }\right) \right] \Leftrightarrow \\ \delta _{\mathbf{I}^{\ast }}^{FB}\lambda _{1}+\left( 1-\delta _{\mathbf{I} ^{\ast }}^{FB}\right) \cdot \pi ^{\ast } &>\delta \cdot \lambda _{1}+\left( 1-\delta \right) \cdot \pi ^{\ast } \end{array} $$

and

$$ \begin{array}{rl} E\left[ \tilde{\pi},\nu _{2}\left( \mathbf{\cdot }\right) \right] & >E\left[ \tilde{\pi},\nu _{2}\left( \mathbf{\cdot }\mid \mathbf{I}^{\ast }\right) \right] \Leftrightarrow \\ \delta \cdot \lambda _{2}+\left( 1-\delta \right) \cdot \pi ^{\ast } &>\delta _{\mathbf{I}^{\ast }}^{FB}\lambda _{2}+\left( 1-\delta _{\mathbf{I} ^{\ast }}^{FB}\right) \cdot \pi ^{\ast } \end{array} $$

which are implied by Eq. 20 under the assumption that $E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\right) \right] =\pi ^{\ast }$. Observe that these inequalities require $\delta \in \left( 0,1\right)$ since $\delta \in \left\{ 0,1\right\}$ would imply $\delta _{\mathbf{I} ^{\ast }}^{FB}=\delta$. As a consequence of $\delta \in \left( 0,1\right)$, we have from the corollary that $\delta _{\mathbf{I}^{\ast }}^{FB}>\delta$ because $\mu \left( \mathbf{I}^{\ast }\right) <1$ so that the above inequalities hold if and only if

$$ \lambda _{1}>\pi ^{\ast }>\lambda _{2}\text{,} $$

which proves the result.

Part (ii). :

By the corollary, the inequality $E\left[ \tilde{\pi} ,\nu _{1}\left( \mathbf{\cdot }\right) \right] \geq E\left[ \tilde{\pi},\nu _{2}\left( \mathbf{\cdot }\right) \right]$ in Eq. 20 holds if and only if λ ₁ ≥ λ ₂. Consider at first agent 1 and rewrite the relevant part in Eq. 20 as

$$ \begin{array}{rcl} E\left[ \tilde{\pi},\nu _{1}^{opt}\left( \mathbf{\cdot }\mid \mathbf{I} ^{\ast }\right) \right] &>&E\left[ \tilde{\pi},\nu _{1}\left( \mathbf{\cdot } \right) \right] \Leftrightarrow \\ \delta _{\mathbf{I}^{\ast }}^{opt}+\left( 1-\delta _{\mathbf{I}^{\ast }}^{opt}\right) \cdot \pi ^{\ast } &>&\delta \cdot \lambda _{1}+\left( 1-\delta \right) \cdot E\left[ \tilde{\pi},\mu \left( \mathbf{\cdot }\right) \right] \end{array} $$

which, under the assumption that $E\left[ \tilde{\pi},\mu \left( \mathbf{ \cdot }\right) \right] =\pi ^{\ast }$ is equivalent to

$$ \delta _{\mathbf{I}^{\ast }}^{opt}+\left( \delta -\delta _{\mathbf{I}^{\ast }}^{opt}\right) \cdot \pi ^{\ast }>\delta \cdot \lambda _{1}. \label{inequal2} $$

(26)

Observe that δ > 0 is a necessary condition for Eq. 26 to hold. In what follows we prove that δ > 0 is also sufficient. Let δ > 0 and consider at first the case that

$$ \delta -\delta _{\mathbf{I}^{\ast }}^{opt}\leq 0\text{.} \label{case3} $$

(27)

Since the l.h.s. of Eq. 26 is then continuously strictly decreasing in $\pi ^{\ast }\in \left( 0,1\right)$, Eq. 26 is satisfied for all π ^∗ if and only if

$$ \begin{array}{rcl} \delta _{\mathbf{I}^{\ast }}^{opt}& \geq &\delta \cdot \lambda _{1}\Leftrightarrow \\ \frac{1}{\delta \lambda _{1}+\left( 1-\delta \right) \mu (\mathbf{I}^{\ast }) }& \geq &1\text{,} \end{array} $$

which is obviously satisfied. This proves the claim for case (27).

Let δ > 0 and consider now the converse case

$$ \delta -\delta _{\mathbf{I}^{\ast }}^{opt}>0\text{.} \label{case4} $$

(28)

Since the l.h.s. of Eq. 26 is then continuously strictly increasing in $\pi ^{\ast }\in \left( 0,1\right)$, Eq. 26 is satisfied for all π ^∗ if and only if

$$ \delta \geq \delta \cdot \lambda _{1}\text{,} $$

which is obviously satisfied. This proves our claim that δ > 0 is a necessary and sufficient condition for Eq. 26 to hold. □